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Chapter 7: Problem 12

A recent article in the Cincinnati Enquirer reported that the mean labor cost to repair a heat pump is \(\$ 90\) with a standard deviation of \$22. Monte's Plumbing and Heating Service completed repairs on two heat pumps this morning. The labor cost for the first was \(\$ 75\) and it was \(\$ 100\) for the second. Assume the distribution of labor costs follows the normal probability distribution. Compute \(z\) values for each and comment on your findings.

Short Answer

Expert verified
The labor costs are within one standard deviation from the mean, indicating moderate variation.

Step by step solution

01

Understanding the Context

We know the mean labor cost, \(\mu = 90\), and the standard deviation, \(\sigma = 22\). We also have the labor costs for two repairs: \(75\) and \(100\). This step involves analyzing our given values related to a normal distribution.
02

Formula for Z-values

The formula to calculate a z-value is \( z = \frac{X - \mu}{\sigma} \), where \(X\) is the observed value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. This formula tells us how many standard deviations away \(X\) is from the mean.
03

Calculate Z-value for the First Repair

For the first heat pump with a cost of \(\$ 75\), substitute \(X = 75\) into the z-value formula: \( z = \frac{75 - 90}{22} \). Simplifying, we get: \( z = \frac{-15}{22} \approx -0.682 \).
04

Calculate Z-value for the Second Repair

For the second heat pump with a cost of \(\$ 100\), substitute \(X = 100\) into the z-value formula: \( z = \frac{100 - 90}{22} \). Simplifying, we get: \( z = \frac{10}{22} \approx 0.455 \).
05

Interpret the Z-values

A \(z\)-value indicates how many standard deviations an observation is from the mean. The first repair's \(z\)-value of \(-0.682\) indicates it is about 0.682 standard deviations below the mean. The second repair's \(z\)-value of \(0.455\) indicates it is about 0.455 standard deviations above the mean. Both values are fairly close to the mean, suggesting moderate deviation in labor costs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
The normal distribution is fundamental in statistics, representing a set of data where most of the observations cluster around the mean. This type of distribution is symmetrical, meaning the left side of the graph is a mirror image of the right side. One of the key characteristics of a normal distribution is its bell-shaped curve, often referred to as the "Gaussian curve."
It's defined by two parameters:
  • The mean (\(\mu\))鈥攊t tells us the central tendency, or where the peak of the curve is located.
  • The standard deviation (\(\sigma\))鈥攊t shows how spread out the data is around the mean.
In our context, labor costs for repairing heat pumps are assumed to be normally distributed with a mean of \(\$90\).This assumption helps us calculate how much individual observations, like costs of specific repairs, differ from the expected mean value.
Using the properties of a normal distribution, we can assign probabilities to different costs by determining where they fall on this curve. This is done through calculations like finding \(z\)-values.
Standard Deviation
Standard deviation is a crucial concept in understanding data variability or how spread out the data points are.It is essentially the average distance of each data point from the mean in a data set. A smaller standard deviation means data points are close to the mean, indicating less variability.Conversely, a larger standard deviation indicates more variability among data points.In the example of labor costs for heat pump repairs, the standard deviation is \(\\(22\).This means that generally, repair costs differ from the average mean cost of \(\\)90\) by about \(\$22\) either above or below.
To determine how unusual a particular repair cost is, we use the standard deviation along with the mean to calculate \(z\)-values.Standard deviation thus allows us to understand how much individual repair costs stray from what is typically expected.
Mean Labor Cost
The mean labor cost in our scenario represents the average cost of repairing a heat pump, which is given as \(\\(90\). The mean is the average value of a data set and is calculated by adding all the values and dividing by the number of values.It provides a central reference point, helping us compare individual repair costs to the typical cost. In practical terms, if the labor cost for most repairs hovers around this mean, it implies consistency in pricing.For individual costs, such as \(\\)75\) and \(\$100\), we consider how these costs relate to the mean using the formula for \(z\)-values. This comparison allows us to determine how much more or less expensive these repairs were compared to the average cost.Understanding these variances assists in identifying pricing patterns and adjusting expectations for future repairs.

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Most popular questions from this chapter

A normal distribution has a mean of 50 and a standard deviation of 4. Determine the value below which 95 percent of the observations will occur.

Many retail stores offer their own credit cards. At the time of the credit application, the customer is given a 10 percent discount on the purchase. The time required for the credit application process follows a uniform distribution with the times ranging from 4 minutes to 10 minutes. a. What is the mean time for the application process? b. What is the standard deviation of the process time? c. What is the likelihood a particular application will take less than 6 minutes? d. What is the likelihood an application will take more than 5 minutes?

A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone's privacy. a. How much toothpaste would you expect to be remaining in the tube? b. What is the standard deviation of the amount remaining in the tube? c. What is the likelihood there is less than 3.0 ounces remaining in the tube? d. What is the probability there is more than 1.5 ounces remaining in the tube?

Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of \(\$ 1,280\) and a standard deviation of \(\$ 420\) per year. a. What fraction of the employees cost more than \(\$ 1,500\) per year for dental expenses? b. What fraction of the employees cost between \(\$ 1,500\) and \(\$ 2,000\) per year? c. Estimate the percent that did not have any dental expense. d. What was the cost for the 10 percent of employees who incurred the highest dental expense?

According to the Internal Revenue Service, the mean tax refund for the year 2006 was \(\$ 2,290\). Assume the standard deviation is \(\$ 650\) and that the amounts refunded follow a normal probability distribution. a. What percent of the refunds are more than \(\$ 3,000 ?\) b. What percent of the refunds are more than \(\$ 3,000\) but less than \(\$ 3,500 ?\) c. What percent of the refunds are more than \(\$ 2,500\) but less than \(\$ 3,500 ?\)
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